Optimization

Summary:
One of the main applications of the derivative is
optimization problems — finding the value of one
quantity that will make another quantity reach its largest or smallest
value, as required. Here’s an overview of the solution techniques.

Steps to Optimization

Write the primary equation, the formula for the quantity
to be optimized. The quantity to be optimized is the
dependent variable, and the other variables are
independent variables. Write down whether the dependent
variable is to be maximized or minimized.

Before you can proceed, the primary equation must contain only one
independent variable and the dependent variable. If you have
extra variables, as usually you do, you must construct one or more
secondary equations that involve the independent variables in the
primary equation. (Usually a problem constraint will lead to a secondary
equation.) Substitute the secondary equations in the
primary equation to eliminate the extra variables.

Write down the feasible domain, the values of the
remaining independent variable that make sense in the problem. Usually
this information will come from the problem or your common-sense
knowledge (such as no negative areas).

Write the answer to the problem. Make sure you have answered the
actual question!

Example: Soda Can

A standard US can of soda (or pop, depending on where you
live) holds 12 fluid ounces or 355 ml. Find the dimensions of a
cylindrical can that will use the least amount of aluminum.

Solution:
The dependent variable is the amount of aluminum. Essentially,
you must minimize the surface area of the cylinder. Write the primary
equation: the surface area is the area of the two ends (each
πr²) plus the area of the side or lateral area

to minimize: A = 2πr² + 2πrh

The primary equation contains two independent variables, r
and h. Can you relate them in some way? Yes, the problem constraint is
that the volume equals 355 ml (or 355 cm³). This gives
the secondary equation:

V = πr²h = 355

Since h occurs once in the primary equation, and as a linear
term, it will be easy to eliminate. Solve the secondary equation for
h:

h = 355 / πr²

Substitute in the primary equation:

A = 2πr² + 2πr·355/πr²
⇒ A = 2πr² + 710/r

There are no obvious constraints on the feasible domain,
except that r and h must both be positive. Since the feasible domain
has no endpoints, you need not check whether they are minima.

To try to find maxima or minima, differentiate:

dA/dr = 4πr − 710/r²

d²A/dr² = 4π + 1420/r³

Find critical numbers where dA/dr = 0 or does not
exist. The derivative does not exist at r = 0; however you
can disregard that because r = 0 is outside the feasible
domain.

4πr − 710/r² = 0
⇒
r = cuberoot(710/4π) ≈ 3.84 cm

Is this a minimum, a maximum, or neither? Since d²A/dr&sup2
is positive for all positive r, the critical point
r = 3.84 cm must be a minimum.

The problem asked for the dimensions of the can with
lowest surface area, which means that you also need the height. To find it,
substitute r = 3.84 in the secondary equation and get
h ≈ 7.67 cm.

Answer: A cylindrical can with volume
355 ml will use the least aluminum if its radius is about
3.84 cm and its height is about 7.67 cm.

Check: V = πr²h =
π(3.84²)(7.67) = 355.3 cm³, the same as the
required volume give or take a little rounding difference.

Because a real soda can is not exactly cylindrical, you can’t
expect perfect agreement with these figures. I measured about
3.2 cm radius and 12.7 cm height on a can of Barq’s root
beer, including the top and bottom extensions.

By the way, you may have noticed that the radius is about
½ the height. In fact, you can prove that the cylinder of a
given fixed volume with the lowest surface area will always have
r = h/2. Instead of setting V = 355 keep V as
a letter and treat it as a constant. You will have
r = cuberoot(V/2π) and substituting
V = πr²h gives r = h/2.

Steps to Finding Minimum or Maximum

Differentiate with respect to the independent
variable. You will need both the first and the second derivatives.

Find the critical numbers, where f′ is 0
or does not exist.

For each critical number (call it c),
evaluate the second derivative f″(c) to find whether you have
a maximum, a minimum, or neither, as follows:

If f″(c) is negative, you have a hilltop and c gives
you a maximum.

If f″(c) is positive, you have a valley and c gives
you a minimum.

If f″(c) is 0 or does not exist, you must use the
First Derivative Test, which is:

If f′(x) goes from positive to negative at x=c, you have a
hilltop (maximum) at x=c.

If f′(x) goes from negative to positive at x=c, you have a
valley (minimum) at x=c.

If f′(x) has the same sign on both sides of x=c, you have
neither a maximum nor a minimum at x=c.

You must also evaluate the original function (primary equation) for
each endpoint of the feasible domain, if it has endpoints.
Compare to the function values at the optimum points from step 3.